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I get B but I am not sure. OA pls? Following is the reasoning: stmt 1: we just have two points on the x-axis. we can draw many lines via these two points and having different slopes. Insuff.

stmt2: we have the intersection point in the first quadrant. Now, draw a line perpendicular from (7,2) to the x-axis. This will be a vertical line. Now select two points on the axis just to right and left of (7,0) on x-axis. Let these points be the x-intercepts of the lines. Now the slope of line with a greater x-intercept say, (7.0001, 0 ) will always be -ve [in order to intersect (7,2)] and the slope of the line with a smaller x-intercept say (6.9999, 0 ) will always be +ve [in order to intersect (7,2)]

Approach of economist is nice. Lets do it more mathematically. 1)insuff. No need to explain. 2)insuff. No need to explain.

together line k=> y=mx+n line j=> y=tx+y we know from 1 that n is greater than y. And we know from 2 that they are intersecting at point (7,2). That makes 2=7m+n 2=7t+y 7m+n=7t+y 7m+n<7t+n 7m<7t m<t

In y = mx + b equation, m = slope and b is the Y intercept. From your represenation, how can we say that

maliyeci wrote:

we know from 1 that n is greater than y.

when variables n and y are the Y intercepts. Not X intercepts.

Can you please explain how you made this Inequality based on n > y(assuming y intercepts of L > y intercept of K). I am confused.

maliyeci wrote:

7m+n<7t+n 7m<7t m<t

I beg your pardon for the intercepts. Yes you are rigth. I gave the Y intercept. The x intercepts are respectively; -n/m and -y/t But they are sufficient to find an answer. That is the solution. 7m+n=2 n=2-7m n/m=2/m-7 same as for the other y/t=2/t-7 since -n/m is greater than -y/t; y/t is greater than n/m so 2/m-7>2/t-7 2/m>2/t so we can find the solution C is suff.

Re: xy–coordinate plane [#permalink]
03 Oct 2009, 12:27

4

This post received KUDOS

Expert's post

Sorry to disappoint you guys, but seems that the answer should be E.

First of all to solve this question think that the best way is graphic approach:

We all agree that (1) and (2) alone are not sufficient.

Lets combine --> draw only X-axis. k intercept is greater than j intercept --> k intercept (k1) point right to j intercept point (j1) --> Interception of lines is above X-axis (point 7;2) --> it will give us only three possible ways to draw the lines: A. j has negative slope, k has negative slope --> slope of k>slope of j, (because j is closer to vertical line); B. j has positive slope, k has negative slope --> slope of k<slope of j; C. j has positive slope, k has positive slope --> slope of k>slope of j, (because k is closer to vertical line);

So (1)+(2) not sufficient. Answer E.

BUT the way you were solving also should get E:

f(k)=mx+n f(j)=tx+p (1) The x-intercept of line k is greater than the x-intercept of line j --> 0=mx+n; 0=tx+p --> -n/m>-p/t or n/m<p/t --> not sufficient; (2) Lines k and j intersect at (7, 2) 2=7m+n; 2=7t+p --> 7m+n=7t+p -->not sufficient;

Combining the way you did: n=2-7m --> n/m=2/m-7 p=2-7t --> p/t=2/t-7 n/m<p/t --> 2/m-7<2/t-7 --> 1/m<1/t --> (t-m)/mt<0 And here is the catch, from above statement you can not determine whether m>t or not. t=1<m=3 statement is true and t=1>m=-3 statement is also true. Answer E. _________________

Re: xy–coordinate plane [#permalink]
03 Oct 2009, 12:31

Bunuel wrote:

Sorry to disappoint you guys, but seems that the answer should be E.

First of all to solve this question think that the best way is graphic approach:

We all agree that (1) and (2) alone are not sufficient.

Lets combine --> draw only X-axis. k intercept is greater than j intercept --> k intercept (k1) point right to j intercept point (j1) --> Interception of lines is above X-axis (point 7;2) --> it will give us only three possible ways to draw the lines: A. j has negative slope, k has negative slope --> slope of k>slope of j, (because j is closer to vertical line); B. j has positive slope, k has negative slope --> slope of k<slope of j; C. j has positive slope, k has positive slope --> slope of k>slope of j, (because k is closer to vertical line);

So (1)+(2) not sufficient. Answer E.

BUT the way you were solving also should get E:

f(k)=mx+n f(j)=tx+p (1) The x-intercept of line k is greater than the x-intercept of line j --> 0=mx+n; 0=tx+p --> -n/m>-p/t or n/m<p/t --> not sufficient; (2) Lines k and j intersect at (7, 2) 2=7m+n; 2=7t+p --> 7m+n=7t+p -->not sufficient;

Combining the way you did: n=2-7m --> n/m=2/m-7 p=2-7t --> p/t=2/t-7 n/m<p/t --> 2/m-7<2/t-7 --> 1/m<1/t --> (t-m)/mt<0 And here is the catch, from above statement you can not determine whether m>t or not. t=1<m=3 statement is true and t=1>m=-3 statement is also true. Answer E.

This exactly why I said IMO E . Thanks for explanation Bunuel!

Re: xy–coordinate plane [#permalink]
22 Nov 2012, 05:22

Bunuel wrote:

Lets combine --> draw only X-axis. k intercept is greater than j intercept --> k intercept (k1) point right to j intercept point (j1) --> Interception of lines is above X-axis (point 7;2) --> it will give us only three possible ways to draw the lines: A. j has negative slope, k has negative slope --> slope of k>slope of j, (because j is closer to vertical line); B. j has positive slope, k has negative slope --> slope of k<slope of j; C. j has positive slope, k has positive slope --> slope of k>slope of j, (because k is closer to vertical line);

Hi Bunuel,

Request you to please look at the colored portion again and confirm if its correct. Isn't in Option A->Slope of j > Slope of K. If not, please provide your reasoning, may be I'm lacking some concept.

Re: xy–coordinate plane [#permalink]
22 Nov 2012, 06:08

1

This post received KUDOS

Expert's post

imhimanshu wrote:

Bunuel wrote:

Lets combine --> draw only X-axis. k intercept is greater than j intercept --> k intercept (k1) point right to j intercept point (j1) --> Interception of lines is above X-axis (point 7;2) --> it will give us only three possible ways to draw the lines: A. j has negative slope, k has negative slope --> slope of k>slope of j, (because j is closer to vertical line); B. j has positive slope, k has negative slope --> slope of k<slope of j; C. j has positive slope, k has positive slope --> slope of k>slope of j, (because k is closer to vertical line);

Hi Bunuel,

Request you to please look at the colored portion again and confirm if its correct. Isn't in Option A->Slope of j > Slope of K. If not, please provide your reasoning, may be I'm lacking some concept.

thanks

A steeper incline indicates a higher absolute value of the slope.

So, if both lines have positive slope, then the line which is steeper has greater slope. If both lines have negative slope, then the line which is steeper has greater absolute value slope, its slope is "more negative", so less than the slope of another line.

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